Given two vectors, a and b, how do I find the Orthogonal Projection?
I've already found the Scalar and Vector Projections.
\begin{align*}
\text{Scalar}&:\quad \frac{-90 + -25 + 24}{\sqrt{9^2+5^2+8^2}};\\
\text{Vector}&:\quad \left(\left(\frac{-91}{\sqrt{170}}\right)\left(\frac{-9}{\sqrt{170}}\right), \left(\frac{-91}{\sqrt{170}}\right)\left(\frac{-5}{\sqrt{170}}\right), \left(\frac{-91}{\sqrt{170}}\right)\left(\frac{8}{\sqrt{170}}\right)\right).
\end{align*}

Your question, as phrased, is nonsensical. Orthogonal projections are with respect to something; I suspect that you want the orthogonal projection onto the plane the two vectors generate. If so, then you need to state that.
–
Arturo MagidinJan 31 '11 at 19:11

2

P.S. Writing things like +-25 is terribly bad form, and prone to errors when reading it later. I suggest you get in the habit of using parentheses, so you would write the numerator of the scalar as -90+(-25)+24.
–
Arturo MagidinJan 31 '11 at 19:27

If that's the case, then yes, I would like to know how to find the orthogonal projection onto the plane the two vectors generate.
–
Math StudentJan 31 '11 at 19:37

3

@Math Student: If what is the case? Again: given that the question, as posed, is nonsensical, you should edit it so that it acquires sense. Don't rely on comments by others to clarify, especially when they (I) are only guessing at what it is you mean. If this is a problem from your textbook, and you are having trouble expressing it in your own words, by all means, provide a direct quotation (you can preface it with a > so that it shows up greyed out).
–
Arturo MagidinJan 31 '11 at 19:39

@Math Student: If you would like to edit your question you can click on the word "edit" in the bottom left corner of your question.
–
Rudy the ReindeerJan 31 '11 at 21:31

1 Answer
1

About the vector projection of $\vec{b}$ onto $\vec{a}$. I am taking its definition from here http://en.wikipedia.org/wiki/Vector_projection where it is given as $(\vec{b} \cdot \vec{e}_a ) \cdot \vec{e}_a$ where $\vec{e}_a$ is the unit vector in $a$-direction.

If the sun shines onto the vectors straight from above, the shadow of $A$ cast onto $B$ is exactly the length of $A$ in the direction of $B$.

The scalar product is defined to be $\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \Theta$ so you know how to calculate this length: $|A| cos \Theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{B}|}$. In your case $\vec{B} = \vec{e}_a$ is a unit vector so its length is one and therefore you get $\vec{b} \cdot \vec{e}_a = |\vec{b}| \cos \Theta$ which is the length of $\vec{b}$ in the direction of $\vec{a}$.

A vector always consists of a length and a direction so now you need to add a direction to the length you just computed. But the direction is the direction of $\vec{a}$. How do you get the direction only without its length? You make it a unit vector like this: $\vec{e}_a = \frac{1}{|\vec{a}|} \vec{a}$.

You see that $length$ $\cdot$ $direction$ is $(\vec{b} \cdot \vec{e}_a ) \cdot \vec{e}_a$ in this case.

Hope this helps. As for the other two cases, scalar projection and orthogonal projection, I don't know what they are. Maybe you could post their definitions given in your lecture here?